I assume you want to put this equation of a parabola in vertex form by completing the square.

First, factor out the 2 from the x terms:

Next, take half the coefficient of x, , square it, , and add it to complete the perfect square trinomial in parentheses. But, you really added twice that amount since we have a factor of outside the parentheses. So, to keep things balanced in the equation, we must subtract twice that amount, , outside the parentheses. 'What one addeth to just one side of an equation, one must taketh away'.

The minimum point is the y-coordinate of the vertex:

The y-intercept can be found from the original function by setting x=0 and solving for y.

The x-intercepts (zeros of the function) are found by setting the original quadratic = 0 and use the quadratic formula to find the values for x.

I assume you want to put this equation of a parabola in vertex form by completing the square.

First, factor out the 2 from the x terms:

Next, take half the coefficient of x, , square it, , and add it to complete the perfect square trinomial in parentheses. But, you really added twice that amount since we have a factor of outside the parentheses. So, to keep things balanced in the equation, we must subtract twice that amount, , outside the parentheses. 'What one addeth to just one side of an equation, one must taketh away'.

so i take it its the same for everything else if you want to complete the square.

thanks for put some info, do you know any good web sites for someone just starting off in algebra